Question

I need the solution for: log(3x) = log 2 + log(x + 4)

Answer 1

`D:\\3x > 0\ and\ x+4 > 0\\x > 0\ and\ x > -4\\therefore\ D:x\in(-4;\ \infty)\\------------------\\log(3x)=log2+log(x+4)\ \ \ |use\ loga+logb=log(a\cdot b)\\\\log(3x)=log[2(x+4)]\iff3x=2(x+4)\\\\3x=2x+8\ \ \ \ |subtract\ 2x\ from\ both\ sides\\\\\boxed{x=8\in D}`

Answer 2

`\hbox{the domain:} \\ 3x>0 \ \land \ x+4>0 \\ x>0 \ \land \ x>-4 \\ x \in (0,+\infty) \\ \\ \log(3x)=\log 2+\log(x+4) \\ 0=\log 2+ \log(x+4) - \log(3x) \\ 0=\log((2 * (x+4))/(3x)) \\ 0=\log((2x+8)/(3x)) \\ 10^0=(2x+8)/(3x) \\ 1=(2x+8)/(3x) \\ 3x=2x+8 \\ 3x-2x=8 \\ \boxed{x=8}`